IR - Graphical Interpretation Sample Question

Question

Hi, 
the OA for the first part is 4,000-5,000. can anyone explain why?
thanks

ganeshiyer1983 · Answer

Hi ,

I am not sure if this is the most elegant way to do this..

Since we want approx answers we can assume that the part of the graph from point A to C is a straight line.

Point A = ( 12,900) Point C = ( 20,5100)

Slope = (5100-900) / ( 20-12) = 525

Hence Equation of line = (y-900) = 525 ( x-12)

Simplifying we get y= 525x - 5400

Now we have to get sum of weights from ages 13 till 19 . Substituting for x we get

525( 13+ 14+ ...19 ) - 7*5400
 525 (112)- 37800 = 21000

From 21 to 30 we can assume average mass as 5300 factoring for a very slight increase from age 20- 30 . Hence total weight is 5300* 10

Thus Total weight = 900+ 21000+ 5100+53000= 80,000

Hence average = 80,000/19 = 4210

Regards

Ganesh

LinaNY · Answer

Thanks Ganesh! Could you please explain why you multiply 5400 by 7? I understand this is a number of all values for x from 13 to 19, but I don't get why you add it to the formula.
thanks

ganeshiyer1983 · Answer

Hi,

The equation is y=525x -5400

for x=13 it becomes y= 525 ( 13) -5400
for x= 14 it becomes y= 525 ( 14)- 5400
for x= 15 it will be y= 525 ( 15) -5400
Similarly continuing till x=19 and we will have 525 ( 19) - 5400

Since we have to find the sum ,when we add all these up there will a total of 7 expressions and each of them will have the term 5400

So we could take 525 as a common factor and we will have 5400 seven times which can be expressed as
525 ( 13+14+15+...19) - (5700 *7)

Does this make sense?

LinaNY · Answer

Yes, it does! thanks!

Rock750 · Answer

Hi,

if you give the graph a closer look, you will see that :

Between 12 and 21, there are 10 values

And between 21 and 30, there are 10 values

So, you can pick 1 value from the former and 1 value from the latter and average the corresponding two values of mass.

For point B (16, 3000) and another point let's call it D ( 25, 5500) , the average mass is : (3000 + 5500)/2= 4250, which is between 4000 and 5000..

This holds true for any values selected from the previous intervals because the graph is incremental

Hope that helps !

LinaNY · Answer

This is what I did in the first place, but I picked A (12, 900) and D (25, 5500) and came up with the average mass of 3200 (900+5500)/2. why didn't it work?
thanks

Rock750 · Answer

Hi LinaNY,
You should take B from the former , it's more representative than any other value picked from the first intervall because it seems to be the middle point i.e the average of the first intervall while for the latter, it seems to have a constant corresponding mass of almost all the values picked from 21 to 30 --> 5500 .
I know it's tricky but when it's about IR, always try to pick the easiest values i.e the most logical ones.
Hope that helps !

MzJavert · Answer

Just by looking at the graph you can see that weights > 5000 account for 1/3 of the age range. With the Median weight at age 16 being 3000 lbs, you can see that the mean will be much higher than 3000 lbs.

See this excellent article for how to determine this without calculations. compare-mean-and-median-in-less-than-20-seconds-146191.html

hellosanthosh2k2 · Answer

Question 1 is test of weighted average concept.

for age 12-19, average weight is roughly 3000 lbs
for age 20-30, average weight is roughly 5500 lbs

3000<------11----------><--------8------->5500
8 (12-20)------------------------------------11(20-30)

weighted average: (8/19) * 2500 + 3000 = (1200)approx + 3000 = 4200 lbs

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IR - Graphical Interpretation Sample Question

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